What is ((y-2)^2)/4+((z-3)^2)/1 =1 ?

The solution to ((y-2)^2)/4+((z-3)^2)/1 =1 is Ellipse with (h,k)=(2,3),a=2,b=1

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((y-2)^2)/4+((z-3)^2)/1 =1

\frac{( y - 2 ) ^{2}}{4} + \frac{( z - 3 ) ^{2}}{1} = 1

(h, k) = (2, 3), a = 2, b = 1

Solution steps

\frac{( y - 2 ) ^{2}}{4} + \frac{( z - 3 ) ^{2}}{1} = 1

Rewrite

\frac{( y - 2 ) ^{2}}{4} + \frac{( z - 3 ) ^{2}}{1} = 1

in the form of the standard ellipse equation

\frac{( y - 2 ) ^{2}}{2 ^{2}} + \frac{( z - 3 ) ^{2}}{1 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (2, 3), a = 2, b = 1

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (2, 3), a = 2, b = 1

\frac{x ^{2}}{- 12} + \frac{y ^{2}}{- 12 - 16} = 1

\frac{( X + 2 ) ^{2}}{9} + \frac{( Y - 1 ) ^{2}}{16} = 1

\frac{x}{x ^{2} + y ^{2}} \leq \frac{1}{2}

\frac{( x + 1 ) ^{2}}{16} + \frac{( y + 1 ) ^{2}}{9} = 1

a^{2} + 2 b^{2} = 1

What is ((y-2)^2)/4+((z-3)^2)/1 =1 ?
The solution to ((y-2)^2)/4+((z-3)^2)/1 =1 is Ellipse with (h,k)=(2,3),a=2,b=1